1 THE GAME. Two players, i=1, 2 U i : concave, strictly increasing f: concave, continuous, f(0) 0 β (0, 1): discount factor, common

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1 1 THE GAME Two players, i=1, 2 U i : concave, strictly increasing f: concave, continuous, f(0) 0 β (0, 1): discount factor, common With Law of motion of the state: Payoff: Histories: Strategies: k t+1 = f(k t ) c 1 t c 2 t β t U i (Ct 1 ), i = 1, 2 t=0 H t = {(c 1 1, c 2 1,..., c 1 t, c 2 t )} σ i t : H t [0, ) A game is naturally defined. Equilibria are Subgame Perfect Equilibria. 1

2 2 ALLOCATION RULE For two consumption rates, c 1, c 2 : c 1, A 1 (c 1, c 2, k) f(k) c 2 A 2 is defined similarly. if c 1 + c 2 f(k), or c 1 f(k) 2 ifc 1 + c 2 f(k) and c 1 f(k) 2 c 2 f(k) if c 1, c 2 f(k) 2 2 2

3 3 FAST CONSUMPTION STRATEGY c i (k) = f(k), i = 1, 2 NOTE: ( c 1, c 2 ) t 0 is a SPE. Let V (k) be the total utility to the ith player of the fast consumption SPE. 3

4 4 THE SECOND BEST PROBLEM Given k, initial stock, the Second Best Value is given by V sb (k) sup t β t [α 1 U 1 (c 1 t ) + α 2 U 2 (c 2 t )] subject to f(k t ) c 1 t c 2 t = k t+1, and k 0 = k, where the supremum is taken over the sequences (c 1 t, c 2 t ) t 0 of outcomes of SPE. THE FIRST BEST (FB) value is defined as the supremum over feasible consumption paths (c 1 t, c 2 t ) t 0, and is denoted V. 4

5 5 TRIGGER STRATEGY PAIR Trigger Strategy Pair is described by: 1. An agreed consumption path, (c 1 t, c 2 t ) t 0 ; 2. The threat to switch to a fast consumption, if a defection of either player from the agreed consumption path is detected. 5

6 6 INDIVIDUAL RATIONALITY CONSTRAINT If (c 1 t, c 2 t ) t 0 is a SPE OUTCOME, then: The DEFECTION VALUE V D β t U i (c i t) V (k) t i (k, c) = max{sup c 1 0U i (c 1 )+βu i ( f(f(k) c c1 ) )+ β2 2 1 β U i( f(0) 2 ), V i (k)} The solution to this problem is denotes as c D i (k, c) Lemma Any SPE outcome (c 1 t, c 2 t ) t 0 is the outcome of a trigger strategy equilibrium. Therefore, the Second Best Problem can be equivalently reformulated as: V sb (k) sup {(c 1 t,c 2 t ) t 0} β t [α 1 U 1 (c 1 t ) + α 2 U 2 (c 2 t )] t subject to f(k t ) c 1 t c 2 t = k t+1, k 0 = k ( ) β n U i (c i t+n) Vi D (k t, c i t ) t = 1, 2,...i = 1, 2. n=0 The constraint (*) is referred to as Incentive Compatibility Constraint. 6

7 7 THE SYMMETRIC CASE U i = U, α i = 1 2, i = 1,2; The Bellman operator for The Second Best Problem: where T V (k) T is not a contraction. max {U(c) + βv (f(k) 2c)} c A(k,ν) A(k, V ) {c : c 0, V (k) V D (k, c)} 7

8 THEOREM Assume lim k f (k)β < 1, then: 1. A solution to the Second Best Problem exists; 2. V sb is upper-semicontinuous. Remark. This is not a standard Dynamic Programming Problem. It is true that: V sb (k) = max {U(c) + βv sb(f(k) 2c)} c A(k,ν) and analogous equations for the non-symmetric case; But even if U, f are concave, V sb is not necessarily concave. The symmetric second-best equilibrium is the greatest growing one. PROPOSITION: For a given k, let (c t 1, c t 2) t 0 be a second best equilibrium outcome, starting from k, which is symmetric and such that V i (k) = V D (k, c i ) (that is second best is not first best), i = 1, 2. If k 1 = f(k) c 1 c 2 and k is the next period capital for any subgame perfect equilibrium, then k k 1. 8

9 8 A FIRST EXAMPLE U i (c) = c1 ɛ 1 ɛ, ɛ (0, 1), α 1 = α 2 = 1/2 y = f(k) = ak, a 1 FIRST BEST SOLUTION ĉ 1 (y) = ĉ 2 (y) = λy λ = 1 2 [1 β 1 ɛ a 1 ɛ ɛ ] where s(λ) λ 1 ɛ (1 ɛ)[1 β(a(1 2λ)) 1 ɛ ] V (k) = s( λ)k 1 ɛ, In general, the value of a consumption policy c(y) = λy for λ (0, 1 2 ) is V (k) = s(λ)k 1 ɛ Defection from a policy c(y) = λy is c D (k, λy) = λ D y, λ D = M(1 λ) where V D (k, λ y ) = s D (λ)k 1 ɛ ; M 1 = 1 + ( aβ 2 ) 1 ɛ s D (λ) = (1 λ)1 ɛ M ɛ 1 ɛ 2 a > 1 9

10 Proposition. The Second Best Consumption policy is given by: 1. c sb (y) = λy if s( λ) s D ( λ) 2. c sb (y) = λ sb y if s( λ) < s D ( λ) where N λ sb min{λ : λ [ λ, 1 + M ] and s(x) = s D(λ)} If a = 3.3, β = 0.325, ɛ = 0.5, then V D (k, ĉ(k)) > V (k) for k > 0 λ = 0.326, λ sb = FIRST BEST: Growth at 15 %, SECOND BEST: Contraction at % Another example If a = 1.058, β = 0.95, ɛ = 0.1, aβ>1 then FIRST BEST: Growth at 5.2 % SECOND BEST: Contraction at -99 % 10

11 In general: 11

12 PROPOSITION Assume 1. for some k, V (k) < V D (k, λk) 2. there exists a c such that: U( c + β ν(f(k) 2 c)) = ν D (k, c), and then, f(k) 2 c k; f(k) 2c sb (k) k 12

13 9 EXAMPLE OF WEALTH DEPENDENT GROWTH Let: a = 1.875; b = 0.2; y = ak + be; β = 0.55; e = 0.1; ɛ = 0.45 FIRST BEST SOLUTION: Growth at 7 % for k > 10 3 But, V (k) V D (k, λy) for k 1 V (k) < V D (k, λy) for k 0.9 for k [0.1, 0.4] the proposition applies, so SECOND BEST SOLUTION: No growth for k [0.1, 0, 4] Another example a = ; b = 0.2β = 0.95; e = 0.2aβ > 1 No growth for k [0.3, 0.9] First best for k

14 10 A CLASSICAL TRAP with Aβ > 1 > b > Bβ A = 5 2 ; β = 1 2 ; B > 2 { Ak if k < 1 f(k) = A + B(k 1) if k 1 { c if c 1 U(c) = 1 + b(c 1) if c 1 14

15 10.1 AN OLSON ECONOMY U(c) = c f(k) = k α FIRST BEST SOLUTION Steady state: k = (αβ) 1 1 α Consumption: { 0 if k (k ĉ(k) = ) 1 α k α k if k (k ) 1 α 2 DEFECTION from FIRST BEST: Consumption: { 0 iff (k c D α c) α 1 2 (k, c) = αβ k α c γ otherwise where γ = ( αβ 2 ) 1 1 α 15

16 11 INCENTIVE COMPATIBLE STEADY STATE k such that: 1 f(k) k 1 β 2 V D (k, f(k) k ) ( ) 2 The set of k s which satisfies (*) is an interval, [k, k]. Two cases: FIRST BEST = SECOND BEST 16

17 Failure of Blackwell with incentive constraints that depend on the control: 1. Standard Case (Discounting condition) T (v) = Max ct 0 U (c t ) + βv (f (k t 2c t )) T (v + α) = Max ct 0 U (c t ) + β (v (f (k t 2c t )) + α) = Max ct 0 U (c t ) + βv (f (k t 2c t )) + βα So Blackwell holds because equality suffi ces: T (v + α) T (v) + βα 17

18 where 2.Case with incentive constaints: T (v) = Max ct R(0) U (c t ) + βv (f (k t 2c t )) R (α) is defined by the set c (k) satisfying R (α) = { c t v(k) + α > v d (k, c) } v d (k, c) k > 0, v d (k, c) c < 0 Note here that vd (k,c) < 0 because deviating from agreed upon consumption c path {c t } is more beneficial when c t is lower-see Benhabib-Rustichini, JOEG,

19 T (v + α) = Max ct R(α) U (c t ) + β (v (f (k t 2c t )) + α) = Max ct R(α) U (c t ) + v (f (k t 2c t )) + βα But we cannot claim, as in the standard case T (v) = Max ct R(0) U (c t ) + βv (f (k t 2c t )) T (v + α) T (v) = Max ct R(α) U (c t ) + v (f (k t 2c t )) because { the set c t R (α) is larger than c t R (0) > 0 This is because ct v(k) + α > v d (k, c) } is larger with α > 0, since v d (k, c) decreases with c. Furthermore, if incentive constraint is binding, you are at the corner: you would like smaller c and accumulate more, but you do not in order to curtail other player from grabbing what you don t consume. Thus when α > 0 constraint is relaxed and you can lower c. Result is Blackwell fails: continuity of v cannot be assured. 19

20 Intuition for the discontinuity: if the value function jumps at k, why not restrain consumption by ε, and take a jump in value by driving k a little higher. Answer, If you were to curtail consumption even just a bit, you create incentives for the other player to grab more. 20